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## Integral Calculus

### Course: Integral Calculus>Unit 1

Lesson 17: Integration by parts

# Integration by parts review

Review your integration by parts skills.

## What is integration by parts?

Integration by parts is a method to find integrals of products:
integral, u, left parenthesis, x, right parenthesis, v, prime, left parenthesis, x, right parenthesis, d, x, equals, u, left parenthesis, x, right parenthesis, v, left parenthesis, x, right parenthesis, minus, integral, u, prime, left parenthesis, x, right parenthesis, v, left parenthesis, x, right parenthesis, d, x
or more compactly:
integral, u, space, d, v, equals, u, v, minus, integral, v, space, d, u
We can use this method, which can be considered as the "reverse product rule," by considering one of the two factors as the derivative of another function.

## Practice set 1: Integration by parts of indefinite integrals

Let's find, for example, the indefinite integral integral, x, cosine, x, d, x. To do that, we let u, equals, x and d, v, equals, cosine, left parenthesis, x, right parenthesis, d, x:
integral, x, cosine, left parenthesis, x, right parenthesis, d, x, equals, integral, u, d, v
u, equals, x means that d, u, equals, d, x.
d, v, equals, cosine, left parenthesis, x, right parenthesis, d, x means that v, equals, sine, left parenthesis, x, right parenthesis.
Now we integrate by parts!
\begin{aligned} \displaystyle\int x\cos(x)\,dx &=\displaystyle\int u\,dv \\\\ &=uv-\displaystyle\int v\,du \\\\ &=\displaystyle x\sin(x)-\int\sin(x)\,dx \\\\ &=x\sin(x)+\cos(x)+C \end{aligned}
Problem 1.1
• Current
integral, x, e, start superscript, 5, x, end superscript, d, x, equals, question mark

Want to try more problems like this? Check out this exercise.

## Practice set 2: Integration by parts of definite integrals

Let's find, for example, the definite integral integral, start subscript, 0, end subscript, start superscript, 5, end superscript, x, e, start superscript, minus, x, end superscript, d, x. To do that, we let u, equals, x and d, v, equals, e, start superscript, minus, x, end superscript, d, x:
u, equals, x means that d, u, equals, d, x.
d, v, equals, e, start superscript, minus, x, end superscript, d, x means that v, equals, minus, e, start superscript, minus, x, end superscript.
Now we integrate by parts:
\begin{aligned} &\phantom{=}\displaystyle\int_0^5 xe^{-x}\,dx \\\\ &=\displaystyle\int_0^5 u\,dv \\\\ &=\Big[uv\Big]_0^5-\displaystyle\int_0^5 v\,du \\\\ &=\displaystyle\Big[ -xe^{-x}\Big]_0^5-\int_0^5-e^{-x}\,dx \\\\ &=\Big[-xe^{-x}-e^{-x}\Big]_0^5 \\\\ &=\Big[-e^{-x}(x+1)\Big]_0^5 \\\\ &=-e^{-5}(6)+e^0(1) \\\\ &=-6e^{-5}+1 \end{aligned}
Problem 2.1
• Current
integral, start subscript, 1, end subscript, start superscript, e, end superscript, x, cubed, natural log, x, space, d, x, equals, question mark